@GEEMIC Yes. Just replace non-strict inequalities with strict ones and change the  axes  option:

restart;
A:=plot(1/x, x=0..3, 0..3, thickness=2):
B:=plots:-inequal({x>0,x<1}, x=0..3,y=0..3, optionsfeasible = [color = "LightBlue"]):
plots:-display(A,B, axes=frame);

@vv  Thank you. If we add up the areas of all triangles, we get for each of the two solutions  x+y+z+1003 = 1922 , 1922+100 = 2022

Everyone can see my solution at the link 
http://math.hashcode.ru/questions/237436/%D0%B3%D0%B5%D0%BE%D0%BC%D0%B5%D1%82%D1%80%D0%B8%D1%8F-%D1%91%D0%BB%D0%BA%D0%B0-%D0%B2-%D1%82%D1%80%D0%B5%D1%83%D0%B3%D0%BE%D0%BB%D1%8C%D0%BD%D0%B8%D0%BA%D0%B0%D1%85

@mehdibaghaee I don't know how you got it, you probably used palettes to enter these expressions. I advise you to use 1Dmath input instead of palettes and 2Dmath input, then such problems will disappear.

@Dkunb  I do not have Maple 2021 (only Maple 2018 and older), so I cannot check how the code works in your version of Maple.

@Dkunb  Strange, in Maple 2018 this code works as expected. Did you do  restart  in the beginning? Your screenshot doesn't show it.

@jud  vv have already answered you that this problem can probably only be solved numerically.

@vv  Thank you. I found the reason for the disappearance of this solution. I missed one condition in ListTools:-Categorize command. If we add it, then the tetrahedron you specified appears. I will now edit my answer.

@vv  The volume of this tetrahedron is 21 (not 210) and it is on my list.

@mmcdara  I use Maple 2018.2

Here's another example with similar behavior:

add(1,2,3); # Error
S:=1,2,3;
add(S); # OK

This is a very simple task. See help on the commands  diff  and  maximize .

@Rouben Rostamian

 The simplification  alpha__max__1  with Maple:

alpha__max__1 := Pi - Pi*sqrt(29 - 4*sqrt(43)*cos(arctan((9*sqrt(191))/1121)/3) - 4*sqrt(3)*sqrt(43)*sin(arctan((9*sqrt(191))/1121)/3))/3;
e:=(1/3)*arctan(9*sqrt(191)*(1/1121)):
thaw(convert(subs(e=freeze(e), alpha__max__1), phaseamp, freeze(e)));

@Randy233  See the update to my solution above.

@srikantha087 It all depends on the size of the matrix, not how it was obtained. For your case I am using the function  i->0.5*i-0.75  that maps the set  {1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5, 10.5, 11.5}  onto the set  {0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0} . I took {1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5, 10.5, 11.5} instead of  {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}  so that the marks are in the middle of the base of the pillars.

@AHSAN  The line

p:=unapply(eval(p(x),%), x);

gives a standard way to define  p  as a function of  x  for later use.

To find  lambda  we only leave one condition  p(0)=0  and then express  lambda  through  k  using the second condition  p(1)=0 :

restart;
h := k - (k - 1)*x;
DE := diff(p(x), x) = 6/h^2 - 12*lambda/h^3;     
BC := p(0) = 0;                    
dsolve({BC, DE}, p(x));
eval(rhs(%),x=1);
solve(%,lambda);